# Proof

(related to Lemma: Subgroups of Cyclic Groups)

Let $$(G,\ast)$$ be a cyclic group generated by $$g$$, i.e. $$G=\langle g \rangle$$ and let $$H\subseteq G$$ be its subgroup. We can easily recognize trivial subgroups $$H=\{e\}=\langle g^0 \rangle$$ and $$H=G=\langle g^1 \rangle$$ as cyclic. Thus, assume $$H$$ being another subgroup of $$G$$. In such a case, $$H$$ must contain at least one element $$x=g^k$$ with a natural number $$k > 1$$. Consider the set of all such natural numbers, more precisely

$K:=\{k~|~x=g^k, x\in H,~ k > 1\}.$

Because $$H$$ is non-empty by assumption, also $$K$$ is non-empty. By virtue of the well-ordering principle, the set $$K$$ contains a smallest element, call it $$b$$. Therefore, also $$H$$ contains the element $$g^b$$. Because $$H$$ is closed under the operation $$\ast$$, it must contain all powers of $$(g^b)^q=g^{b\cdot q}$$, $$q\in\mathbb Z$$ but no other powers of $$g$$. For if it contained a power of $$g$$, e.g. $$g^a$$, whose exponent $$a$$ was not a multiple of $$b$$, then we could use a unique division with a quotient and remainder to write $a=q\cdot b+r$ with $$0 < r < b$$. In this case, however, it would follow that $$g^r\in H$$, and so that $$r\in K$$, in contradiction to the minimality of $$b$$ in $$K$$. Therefore, $$H$$ is cyclic and generated by $$g^b$$, i.e. $$H=\langle g^b \rangle$$.

Github: ### References

#### Bibliography

1. Modler, Florian; Kreh, Martin: "Tutorium Algebra", Springer Spektrum, 2013